Question

An inductor of inductance \(2 μ\text{H}\) is connected in series with a resistance, a variable capacitor and an a.c. source of 10 kHz. The value of capacitance for which maximum current is drawn in to the circuit is \(\frac{1}{x} \text{F}\), where the value of x is (Take \(π^2 = 10\))

• A. \(8000\)
• B. \(600\)
• C. \(400\)
• D. \(1600\)

Hint:

Maximum current needs resonance, so the frequency equals 1 over 2 pi root LC.

Answer 1

The Correct Option is A

Step 1: Use angular frequency
$\omega = 2\pi\times10^4$ and resonance gives $\omega^2LC = 1$.

Step 2: Evaluate
$\omega^2 = 4\pi^2\times10^8 = 4\times10\times10^8 = 4\times10^9$. Then $C = \frac{1}{4\times10^9\times2\times10^{-6}} = \frac1{8000}$. Option (A).

Final Answer:
8000. \[ \boxed{\text{(A)}\ 8000} \]